If it were the momentum or velocity the mean would be zero, but it wouldn't be
exponential. If you just considered the speed (absolute magnitude of velocity) in a
particular direction you get an exponential distribution. Is that what the graph represents?
Brent
On 11/20/2014 5:03 PM, LizR wrote:
The average kinetic energy of an air molecule is zero, I imagine, because they're all
travelling in different directions and cancel out? Or doesn't it work like that?
On 21 November 2014 13:09, meekerdb <meeke...@verizon.net <mailto:meeke...@verizon.net>>
wrote:
On 11/20/2014 3:57 PM, George wrote:
Thanks Bruno, Liz and Richard for your responses.
The topic is extremely controversial… It took me a few months of sleepless
nights
to come to term with these ideas…. but let reason prevail. I am looking
forward to
an open and rational discussion… a background in statistical thermodynamics
would
be helpful.
Bruno, you may not be able to download my pdf file because your Adobe
Reader is not
up to date. If you wish I could simply attach these files to my email.
Please let
me know. You asked me to summarize my post. The best way is with pictures
taken
from my
paper#2.https://sites.google.com/a/entropicpower.com/entropicpower-com/Thermoelectric_Adiabatic_Effects_Due_to_Non-Maxwellian_Carrier_Distribution.pdf?attredirects=0&d=1
(Currently under review)
Figure 7 shows what happens to the energy distribution of a Maxwellian gas
(e.g.
air) as molecules rise from ground level (red) to a given altitude (blue).
Kinetic
energy is converted to potential energy and the distribution shifts to the
left.
Am I correctly interpreting this curve as showing that the kinetic energy
probability density function is an exponential; so that the most probable
kinetic
energy for an air molecule is zero?? Why isn't it the Maxwell-Boltzmann
distribution?
Brent
However when the distribution is renormalized as shown in Figure 8, the
original
distribution is recovered, implying that the gas is isothermal with
elevation. The
Second Law is upheld and Loschmidt is proven wrong….. but only with respect
to
Maxwellian gases.
Now see what happens when a Fermi-Dirac gas (carriers in a semiconductor) is
subjected to a force field as shown in Figure 9. The distribution is
shifted to the
left as elevation increases. However, renormalization does not recover the
original
distribution because it is not exponential. The lower elevation has a higher
temperature than the higher elevation. The Second Law is broken. This
effect can
only be observed in high quality thermoelectric materials (Caltech
experiment).
I have made this calculator program and a simulator publicly available at
my web site.
*Figure. 7.* Un-normalized Maxwell distribution at ground (red/thick) and at
non-zero elevation (blue/thin) showing a shift to a lower kinetic energy, a
drop in
density and a drop in temperature.
*Figure. 8.* Renormalized shifted Maxwell distribution at non-zero elevation
(blue/thin) is identical to original non-shifted distribution at ground
level
(red/thick).
**
*Figure. 11.* Un-normalized Fermi-Dirac distribution at ground (red/thick)
and
non-zero elevation (blue/thin) showing drop in density and drop in
temperature.
*Figure. 12.* Renormalized Fermi-Dirac distributions at ground level
(red/thick)
and at elevation (blue/thin) are different. Elevation lowers energy and
temperature
of gas.
Please look on the right of the pictures for the temperatures at the
ceiling and at
the floor.
George Levy
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